Question: A particle moves along the $x$ -axis. The function $x(t)$ gives the particle's position at any time $t\geq 0$ : $x(t)=t^4-2t^3+3t^2$ What is the particle's velocity $v(t)$ at $t=1$ ? $v(1)=$
Explanation: We have a function for the particle's position, and we need to find the particle's velocity. Since velocity is the rate of change of position, we need to find the derivative of $x(t)$. In other words, if $v(t)$ gives the particle's velocity at any time $t\geq 0$, then $v(t)=x'(t)$. Let's differentiate $x(t)$ to find $v(t)$ : $\begin{aligned} v(t)&=x'(t) \\\\ &=\dfrac{d}{dt}[t^4-2t^3+3t^2] \\\\ &=4t^3-6t^2+6t \end{aligned}$ To find the particle's velocity at $t=1$, we need to evaluate $v(1)$. $\begin{aligned} v({1})&=4({1})^3-6({1})^2+6({1}) \\\\ &=4 \end{aligned}$ Since the velocity is positive, we know the particle is moving to the right. The particle's velocity at $t=1$ is $4$. At $t=1$, the particle is moving to the right.